Chain Rule (HSC SSCE Mathematics Advanced): Revision Notes

Example 1: Identifying inner and outer functions

Question: Consider the function $y = (4x^3 - 4x^2 - 5x - 7)^6$.

If $y = f(g(x))$, identify the inner function $u = g(x)$ and the outer function $f(u)$.

Solution:

Recognise that $(4x^3 - 4x^2 - 5x - 7)^6$ is the result of raising an expression (the inner function) to a power (defined by the outer function).

The given function is $y = (4x^3 - 4x^2 - 5x - 7)^6$.

The inner function $u = g(x)$ is the base of the power:

$u = 4x^3 - 4x^2 - 5x - 7$

The outer function $y = f(u)$ describes what is done to $u$. In this case, $u$ is raised to the power of 6:

$f(u) = u^6$

Remember: A composite function consists of an inner function and an outer function. If $y = f(g(x))$, then $g(x)$ is the inner function (often denoted $u$) and $f(u)$ is the outer function.

Example 2: Using the substitution method

Question: Differentiate $y = (5 - x^2)^3$ using the substitution method.

Solution:

First, identify the inner function $u$ and the outer function $y$ in terms of $u$. Then determine $\frac{dy}{du}$ and $\frac{du}{dx}$, and finally apply the chain rule formula.

Step 1: Set up the substitution

Let the inner function be $u = 5 - x^2$. Then the outer function is $y = u^3$.

Step 2: Find $\frac{du}{dx}$

$u = 5 - x^2$

$\frac{du}{dx} = -2x$

Step 3: Find $\frac{dy}{du}$

$y = u^3$

$\frac{dy}{du} = 3u^2$

Step 4: Apply the chain rule

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

$= 3u^2 \times (-2x)$

$= 3(5 - x^2)^2 \times (-2x)$

$= -6x(5 - x^2)^2$

Note: In the final step, we substitute back $u = 5 - x^2$ to express the answer in terms of $x$.

Example 3: Using the generalised power rule

Question: Differentiate $y = \sqrt{2x^4 + 6}$ in surd form using the generalised power rule.

Solution:

First, rewrite the function in index form, then apply the generalised power rule.

Step 1: Convert to index form

$y = \sqrt{2x^4 + 6}$

$y = (2x^4 + 6)^{\frac{1}{2}}$

Step 2: Apply the generalised power rule

$\frac{dy}{dx} = \frac{1}{2}(2x^4 + 6)^{\frac{1}{2} - 1} \times \frac{d}{dx}(2x^4 + 6)$

$= \frac{1}{2}(2x^4 + 6)^{-\frac{1}{2}} \times (8x^3)$

$= 4x^3(2x^4 + 6)^{-\frac{1}{2}}$

Step 3: Express in surd form

This can also be written with a positive index as:

$\frac{dy}{dx} = \frac{4x^3}{(2x^4 + 6)^{\frac{1}{2}}}$

Or in surd form:

$\frac{dy}{dx} = \frac{4x^3}{\sqrt{2x^4 + 6}}$

Exam tip: When differentiating surds, always convert them to index form first. Remember that $\sqrt{f(x)} = [f(x)]^{\frac{1}{2}}$.